How Market Impact and Dealer Hedging Can Amplify a Gamma Squeeze
Summary
The document derives how delta hedging an options book can affect the underlying when hedge trades move its price. It starts with the delta-hedged profit and loss formula, then models price impact as proportional to the dollar demand created by rebalancing. Under this assumption, the impact changes the underlying’s effective volatility according to the book’s gamma and the impact coefficient. A Black–Scholes pricing assumption then links theta and gamma, allowing the author to express hedged profit and loss in terms of the difference between modeled and effective volatility.
A numerical illustration considers a short at-the-money call and estimates the effect of a small assumed impact over one day. It shows how short gamma hedging can contribute to upward buying pressure and create hedging losses under the chosen assumptions. The treatment is stylized: it assumes continuous hedging, a particular linear impact rule, and Black–Scholes risk estimates. The example is an approximation, not empirical evidence about a specific market squeeze.
Key ideas
- Delta hedging removes the first-order exposure to price changes while leaving gamma exposure in the portfolio.
- When hedge trades move the underlying, gamma-dependent rebalancing can alter the price process’s effective volatility.
- The model’s hedged profit and loss depends on the relationship between realized volatility and the volatility used to price options.
- A short call requires buying stock as its delta exposure changes, which can add upward demand in the assumed impact model.
- The numerical example illustrates the model’s mechanics but does not establish how often or strongly such effects occur in real markets.
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# Gamma squeeze - mathematical explanation
# Gamma squeeze - mathematical explanation
I am trying to understand from a mathematical and financial point of view the mechanism behind the so-called gamma squeeze. Is there a good source to read about this/ My questions are:
- what are the conditions that lead to gamma squeeze?
- what long/short positions favour the creating of a gamma-squeeze situation?
- what are notable examples of such situations?
- and, most importantly, as this situation involves options, I would like to know if option hedging favours or aggravates gamma squeeze and if so, how?
## Answer by Pontus Hultkrantz (score 4)
https://quant.stackexchange.com/a/71261
Your questions on gamma squeeze are nicely answered here and/or here.
We will assume that the risk-free rate $r$ is zero.
Assume that you hold a book $V(t,S)$ consisting of options on the underlying $S$. Itô expanding (Taylor series) $V$, we have that a small pnl change can be written as
\begin{align} dV(t,S) &= \frac{\partial V}{\partial t}dt + \frac{\partial V}{\partial S}dS + \frac{1}{2}\frac{\partial^2 V}{\partial S^2}(dS)^2 + \mathcal{O}(...) \\ &=: V_tdt + V_{s}dS + \frac{1}{2}V_{ss}dS^2. \tag{1} \end{align}
For instance, if a long call option $C$ has theta risk $\theta$, delta risk $\Delta$, and gamma risk $\Gamma$, and we hold a short Call in our book, then $V = -C$ and so $V_t=-\theta$, $V_s=-\Delta$, $V_{ss}=-\Gamma$, i.e. we are short delta and gamma.
Now form the hedged portfolio $P(t,S)$ by continuously hedging our delta by buying and selling the underlying stock. Our pnl dynamics is then \begin{align} dP(t,S) = dV(t,S) - V_s dS = V_t dt +\frac{1}{2}V_{ss}dS^2 \tag{2} \end{align}
Assume now that the true price process follows a generic Itô process $$dS = \alpha(t,S)dt + \beta(t,S)dW \tag{3}.$$
Inserting into equation (2) yields \begin{align} dP(t,S) = \left(V_t +\frac{1}{2}V_{ss}\beta(t,S)^2 \right)dt \tag{4} \end{align}
The pnl from time 0 to T is then $$PnL(0,T) = \int_{t=0}^T dP(t,S).$$
The only assumption we made here is that $dS$ follows a generic Ito process.
##### The Gamma squeeze
The dynamics of a Gamma squeeze is that if we sell an option and then hedge it by buying stocks, we will increase the buy demand and push the price up. We say that our hedging activity incur market impact. If our delta from the options changes by $dV_s$, we need to rebalance our hedges, the number of stocks, by $-dV_s$. If we assume that the market impact is proportional to the demand in dollars, the real stock dynamics can be written as
$$dS = -\lambda S dV_s + \sigma S dW, \tag{5}$$ for some $0 \leq \lambda \ll 1$, i.e. the purchase of one stock to hedge $dV_s=-1$, will move the price up by $\lambda \cdot S$. For instance, $\lambda=1 bp=0.0001$ and if $S=100$, price will move 1 cent. If $\lambda=0$ we have no market impact and the dynamics for $S$ reduces to those under BS.
Itô expanding $dV_s = V_{st}dt + V_{ss}dS + \frac{1}{2}V_{sss}dS^2$ and inserting into eq. (5) and using that $dS$ is a general Ito process by ansatz (3), we find that \begin{align} \alpha(t,S) &= -\lambda\cdot (V_{st} + ½V_{sss}\beta(t,S)^2 )/(1+\lambda SV_{ss}) \\ \beta(t,S) &= \sigma S/(1+\lambda SV_{ss}), \end{align} such that (5) can be simplified into $$dS = \alpha dt + \beta dW = \mathcal{O}(dt) + \frac{1}{1+\lambda S(t) V_{ss}}\sigma S dW \tag{6}.$$
Note that the inclusion of market impact changes the volatility of the stock process. E.g. using $\lambda=1bp$, $S=100$, and $V_{ss}=-20$ we have that the instantaneous vol is around 25% increased.
Inserting the true dynamics for $dS$ from (6) into our pnl in (4), our pnl dynamics can now be written as \begin{align} \boxed{ dP(t,S) = \left(V_t +\frac{1}{2} \sigma^2S^2 V_{ss} \frac{1}{(1+\lambda S V_{ss})^2}\right)dt}. \tag{7} \end{align}
So far we have not assumed any model for pricing and calculating risks of our options, i.e. we need to know the values of our theta $V_t$, delta $V_s$, and gamma $V_{ss}$. Assume that we price and risk under the Black-Scholes (BS) Framework the risks satisfies the BS PDE
$$V_t + \frac{1}{2}\sigma_{BS}^2S^2 V_{ss} = 0 \tag{8}.$$ That is, there is a direct relationship between our theta and gamma risks. Note that in the BS model it has been implicitly assumed that $dS=\sigma_{BS}S dW$. We already know that is incorrect since the real dynamics is in fact given by (6), so this will cause a hedging error due to the wrong model assumptions for pricing and risking options. However, substituting (8) into (7) to express theta in terms of gamma, we get
\begin{align} \boxed{ dP(t,S) = \frac{1}{2}S(t)^2 V_{ss}(t,S)\left(\frac{\sigma(t)^2}{(1+\lambda S(t) V_{ss}(t,S))^2} - \sigma_{BS}^2\right)dt},\tag{9} \end{align} where $\sigma$ is the real (possibly time-dependent) volatility, and $\sigma_{BS}$ is the constant volatility in the BS model at which we buy and sell options.
In general, for our options book, priced under BS, to be perfectly hedged, we need $dP=0$ implying $$\sigma_{BS} = \sigma(t) \cdot |1+\lambda S(t)V_{ss}(t,S)|\tag{10},$$ which is equivalent of saying that our BS risk model must coincide with the real world dynamics in (5) or (6). Note that if the real world dynamics does not have any market impact, $\lambda=0$, we must have $\sigma_{BS}=\sigma$ to align.
##### Practical Example
If we assume that we set $\sigma_{BS}=\sigma$ when we constructed our pricing model, and enabling market impact our pnl over $[t_0,t_0+dt[$ is
\begin{align} dP(t,S) &= \frac{1}{2}\sigma^2 S(t)^2 V_{ss}(t,S)\left(\frac{1}{(1+\lambda S(t)V_{ss}(t,S))^2} - 1\right)dt,\tag{11} \end{align} Notice that $PnL(0,T):=\int_{t=0}^TdP(t,S)$ is path dependent since $S$ and $V_{ss}(t,S(t))$ both depend on time.
Now assume that we are short one ATM call option with $S=100$, $K=100$, maturity one week $T=1/52$ at vol $\sigma=0.2$. Under BS, this short option has gamma $V_{ss}(t_0,S_0)=-0.144$. Furthermore assume that the market impact is $\lambda =1bp$, i.e. buying one stock pushes the price up by $S\cdot \lambda=100 \cdot 1/10000=0.01$.
Since the option is ATM, we have that $V_{sss}(t,S)\approx 0$ and so gamma $V_{ss}(t,S)$ is rather insensitive to a changes in $S$ around the strike. Furthermore, assume that we are interesting in the pnl over one day, $dt=1/250$, hence $V_{ss}$ does not change much within this day. We approximate $S(t)=S(0)=100$. We get
$$dP(t,S) = -28.77 \cdot(1.002883 - 1)dt = -0.083 \cdot dt,$$ and so $PnL(\text{day 1}) \approx -0.00033 \cdot $
Note that this is the pnl from being short one single option.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.